Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 4 Constructions of Triangles.

## Practice Set 4.1 Geometry 9th Std Maths Part 2 Answers Chapter 4 Constructions of Triangles

Question 1.

Construct APQR, in which QR = 4.2 cm, m∠Q = 40° and PQ + PR = 8.5 cm.

Solution:

As shown in the rough figure draw seg QR = 4.2 cm

Draw a ray QT making an angle of 40° with QR

Take a point S on ray QT, such that QS = 8.5 cm

Now, QP + PS = QS [Q-P-S]

∴ QP + PS = 8.5 cm …….(i)

Also, PQ + PR = 8.5 cm ……(ii) [Given]

∴ QP + PS = PQ + PR [From (i) and (ii)]

∴ PS = PR

∴ Point P is on the perpendicular bisector of seg SR

∴ The point of intersection of ray QT and perpendicular bisector of seg SR is point P.

Steps of construction:

i. Draw seg QR of length 4.2 cm.

ii. Djraw ray QT, such that ∠RQT = 40°.

iii. Mark point S on ray QT such that l(QS) = 8.5 cm.

iv. Join points R and S.

v. Draw perpendicular bisector of seg RS intersecting ray QT.

Name the point as P.

vi. Join the points P and R.

Hence, ∆PQR is the required triangle.

Question 2.

Construct ∆XYZ, in which YZ = 6 cm, XY + XZ = 9 cm, ∠XYZ = 50°.

Solution:

As shown in the rough figure draw seg YZ = 6 cm

Draw a ray YT making an angle of 50° with YZ

Take a point W on ray YT, such that YW = 9 cm

Now, YX + XW = YW [Y-X-W]

∴ YX + XW = 9 cm ….(i)

Also, XY + XZ = 9 cm ….(ii) [Given]

∴ YX + XW = XY + XZ [From (i) and (ii) ]

∴ XW = XZ

∴ Point X is on the perpendicular bisector of seg WZ

∴ The point of intersection of ray YT and perpendicular bisector of seg WZ is j point X.

Steps of construction:

i. Draw seg YZ of length 6 cm.

ii. Draw ray YT, such that ∠ZYT = 50°.

iii. Mark point W on ray YT such that l(YW) = 9 cm.

iv. Join points W and Z.

v. Draw perpendicular bisector of seg WZ intersecting ray YT. Name the point as X.

vi. Join the points X and Z.

Hence, ∆XYZ is the required triangle.

Question 3.

Construct ∆ABC, in which BC = 6.2 cm, ∠ACB = 50°, AB + AC = 9.8 cm.

Solution:

As shown in the rough figure draw seg CB = 6.2 cm

Draw a ray CT making an angle of 50° with CB

Take a point D on ray CT, such that

CD = 9.8 cm

Now, CA + AD = CD [C-A-D]

∴ CA + AD = 9.8 cm …….(i)

Also, AB + AC = 9.8 cm ……(ii) [Given]

∴ CA + AD = AB + AC [From (i) and (ii)]

∴ AD = AB

∴ Point A is on the perpendicular bisector of seg DB

∴ The point of intersection of ray CT and perpendicular bisector of seg DB is point A.

Steps of construction:

i. Draw seg BC of length 6.2 cm.

ii. Draw ray CT, such that ∠BCT = 50°.

iii. Mark point D on ray CT such that l(CD) = 9.8 cm.

iv. Join points D and B.

v. Draw perpendicular bisector of seg DB intersecting ray CT. Name the point as A.

vi. Join the points A and B.

Hence, ∆ABC is the required triangle.

Question 4.

Construct ∆ABC, in which BC = 3.2 cm, ∠ACB = 45° Solution:and perimeter of AABC is 10 cm.

Solution:

Perimeter of ∆ABC = AB + BC + AC

∴ 10 = AB + 3.2 + AC

∴ AB + AC = 10 – 3.2

∴ AB + AC = 6.8 cm

Now, In ∆ABC

BC = 3.2 cm, ∠ACB = 45° and AB + AC = 6.8 cm ….(i)

As shown in the rough figure draw j seg BC = 3.2 cm

Draw a ray CT making an angle of 45° with CB

Take a point D on ray CT, such that

CD = 6.8 cm

Now, CA + AD = CD [C-A-D]

∴ CA + AD = 6.8 cm …(ii)

Also, AB + AC = 6.8 cm ….(iii) [From (i)]

∴ CA + AD = AB + AC [From (ii) and (iii)]

∴ AD = AB

∴ Point A is on the perpendicular bisector of seg DB

∴ The point of intersection of ray CT and perpendicular bisector of seg DB is point A.

Steps of construction:

i. Draw seg BC of length 3.2 cm.

ii. Draw ray CT, such that ∠BCT = 45°.

iii. Mark point D on ray CT such l(CD) = 6.8 cm. that

iv. Join points D and B.

V. Draw perpendicular bisector of seg DB intersecting ray CT. Name the point as A.

vi. Join the points A and B.

Hence, ∆ABC is the required triangle.